${\int }_C\mathbf{F}·\mathbf{dr}$

$\={\int }_a^{b}f \mathrm{dx}\+g \mathrm{dy}$

$\={\int }_0^{2 \mathrm{\π}}\left[\left(\genfrac{}{}{0ex}{}{x y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}x\left(t\right)\+\left(\genfrac{}{}{0ex}{}{x\+y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=x\left(t\right)\,y\=y\left(t\right)}\right)\cdot \frac{d}{\mathrm{dt}}y\left(t\right)\right] \mathrm{dt}$

$\={\int }_0^{\arctan \left(2\right)}\left[\left(3\+2 \cos \left(t\right)\right)\left(4\+\sin \left(t\right)\right)\cdot \left(-2 \sin \left(t\right)\right)\+\left(\left(3\+2 \cos \left(t\right)\right)\+\left(4\+\sin \left(t\right)\right)\right)\cdot \cos \left(t\right)\right] \mathrm{dt}$

$\={\int }_0^{\arctan \left(2\right)}\left(4{\cos \left(t\right)}^3-15\cos \left(t\right)\sin \left(t\right)\+8{\cos \left(t\right)}^2-24\sin \left(t\right)\+3\cos \left(t\right)-6\right) \mathrm{dt}$

$\=\frac{538}{75}\sqrt{5}-\frac{142}{5}-2 \arctan \left(2\right)$ ≐ $-14.57$

Initialize

Loading Student:-VectorCalculus

Define the vector field F

$⟨x y\,x\+y⟩$ = $\left[\begin{array}{c}xy\\ x+y\end{array}\right]$$\stackrel{\text{to Vector Field}}{\to }$$\left[\begin{array}{c}xy\\ x+y\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$F$$$

Obtain the line integral via the Context Panel

$\mathbf{F}$

$\stackrel{\text{line integral}}{\to }$

${\∫}_0^{\arctan \left(2\right)}\left(-2\left(2\cos \left(t\right)\+3\right)\left(4\+\sin \left(t\right)\right)\sin \left(t\right)\+\left(2\cos \left(t\right)\+7\+\sin \left(t\right)\right)\cos \left(t\right)\right)\ⅆt$

$\=$

$-\frac{142}{5}\+\frac{538}{75}\sqrt{5}-2\arctan \left(2\right)$

$\stackrel{\text{at 10 digits}}{\to }$

$-14.57423649$

Define the path parametrically, as a position vector

$\mathbf{R}\=⟨3\+2 \cos \left(t\right)\,4\+\sin \left(t\right)⟩$$\stackrel{\text{assign}}{\to }$

Obtain T, a unit tangent vector along the path

$\mathbf{R}$ = $\stackrel{\text{tangent vector}}{\to }$ $\stackrel{\text{Euclidean-normalize}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$T$$$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$∥\frac{\ⅆ}{\ⅆ t} \mathbf{R}∥$ = $\sqrt{-3{\cos \left(t\right)}^2\+4}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

Obtain the integrand $\mathbf{F}·\mathbf{T} \mathrm{ds}$ 

$\mathbf{F}·\mathbf{T} \mathrm{\ρ}$

$\left(-\frac{2\left(3\+2\cos \left(t\right)\right)\left(4\+\sin \left(t\right)\right)\sin \left(t\right)}{\sqrt{-3{\cos \left(t\right)}^2\+4}}\+\frac{\left(7\+2\cos \left(t\right)\+\sin \left(t\right)\right)\cos \left(t\right)}{\sqrt{-3{\cos \left(t\right)}^2\+4}}\right)\sqrt{-3{\cos \left(t\right)}^2\+4}$

$\stackrel{\text{simplify}}{\=}$

$4{\cos \left(t\right)}^3-15\cos \left(t\right)\sin \left(t\right)\+8{\cos \left(t\right)}^2-24\sin \left(t\right)\+3\cos \left(t\right)-6$

$\stackrel{\text{integrate w.r.t. t}}{\to }$

${\∫}_0^{2\mathrm{\π}}\left(4{\cos \left(t\right)}^3-15\cos \left(t\right)\sin \left(t\right)\+8{\cos \left(t\right)}^2-24\sin \left(t\right)\+3\cos \left(t\right)-6\right)\ⅆt$

$\=$

$-4\mathrm{\π}$

Initialize

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x y\,x\+y⟩\right)\:$

Apply the LineInt command

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Arc}\left(\mathrm{Ellipse}\left(⟨3\,4⟩\,2\,1\,0\right)\,0\,\mathrm{\pi }\/4\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{\arctan \left(2\right)}\left(-2\left(2\cos \left(t\right)\+3\right)\left(4\+\sin \left(t\right)\right)\sin \left(t\right)\+\left(2\cos \left(t\right)\+7\+\sin \left(t\right)\right)\cos \left(t\right)\right)\ⅆt$

$\mathrm{LineInt}\left(\mathbf{F}\,\mathrm{Arc}\left(\mathrm{Ellipse}\left(⟨3\,4⟩\,2\,1\,0\right)\,0\,\mathrm{\pi }\/4\right)\right)$ = $-\frac{142}{5}\+\frac{538}{75}\sqrt{5}-2\arctan \left(2\right)$$$

Define the path parametrically, as the position vector R

$\mathbf{R}≔⟨3\+2 \cos \left(t\right)\,4\+ \sin \left(t\right)⟩$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$\mathrm{\ρ}≔\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,t\right)\right)$

$\sqrt{-3{\cos \left(t\right)}^2\+4}$

Obtain T, a unit tangent vector along the path

$\mathbf{T}≔\mathrm{TangentVector}\left(\mathbf{R}\,t\,\mathrm{normalized}\right)$

Form and evaluate the requisite line integral

$\mathrm{Int}\left(\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{T}\right)\cdot \mathrm{\ρ}\right)\,t\=0..\arctan \left(2\right)\right)\=\mathrm{int}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{T}\right)\cdot \mathrm{\ρ}\,t\=0..\arctan \left(2\right)\right)$

${\∫}_0^{\arctan \left(2\right)}\left(4{\cos \left(t\right)}^3-15\cos \left(t\right)\sin \left(t\right)\+8{\cos \left(t\right)}^2-24\sin \left(t\right)\+3\cos \left(t\right)-6\right)\ⅆt\=-\frac{142}{5}\+\frac{538}{75}\sqrt{5}-2\arctan \left(2\right)$